NOTES ON INFINITE SEQUENCES AND SERIESmlerma/courses/b17-99f/seq.pdf · NOTES ON INFINITE SEQUENCES AND SERIES 5 2.3. Telescopic Series. Telescopic series areseries forwhich allterms

NOTES ONINFINITE SEQUENCES AND SERIES

MIGUEL A. LERMA

1. Sequences

1.1. Sequences. An infinite sequence of real numbers is an orderedunending list of real numbers. E.g.:

1, 2, 3, 4, . . .

We represent a generic sequence as a1, a2, a3, . . . , and its n-th as an.In order to define a sequence we must give enough information to

find its n-th term. Two ways of doing this are:

1. With a formula. E.g.:

an =1

n

an =1

10n

an =√

3n− 7

2. With a recursive definition. E.g.: the Fibonacci sequence 1, 1, 2, 3, 5, 8, . . . ,in which each term is the sum of the two previous terms:

F1 = 1F2 = 1

Fn+1 = Fn + Fn−1

1.2. Limit of a Sequence. We say that a sequence an converges toa limit L if the difference |an−L| can be made as small as we wish bytaking n large enough. We write an → L, or more formally:

limn→∞

an = L .

E.g.:

limn→∞

1

n= 0 .

Date: 10/12/1999.1

2 MIGUEL A. LERMA

If a sequence does not converge we say that it diverges. E.g., thefollowing sequences diverge:

n = 1, 2, 3, 4, · · · → diverges (to +∞)

(−1)n = −1, 1,−1, 1, · · · → diverges

1.3. Limit Laws for Sequences. If limn→∞

an = A and limn→∞

bn = B,

then:

limn→∞

(an + bn) = A + B

limn→∞

(an − bn) = A− B

limn→∞

(an bn) = A B

limn→∞

(an/bn) = A/B (provided B 6= 0)

So, a “complicated” limit such as L = limn→

1 + 1n

3 + 110n

can be computed

by replacing smaller parts of it with their limits 1/n → 0, 1/10n → 0:

L =1 + 0

3 + 0=

1

3.

1.4. Squeeze Law. If an ≤ cn ≤ bn, and an and bn have the samelimit: an → L, bn → L, then cn has also the same limit: cn → L. Thiscan be used to compute limits such as the following one:

limn→∞

sin n

n.

In this case we have:

−1

n≤ sin n

n≤ 1

n.

Since −1/n → 0 and 1/n → 0 thensin n

n→ 0 also.

1.5. Limits of Functions of Sequences. If an = f(n) for some func-tion f and lim

x→nf(x) = L, then lim

n→∞an = L. This basically allows us to

replace limits of sequences with limits of functions. In particular this isuseful for using L’Hopital’s rule in computing limits of sequences. E.g:

limn→∞

en

n= lim

x→∞ex

x= (L’Hopital’s rule) = lim

x→∞ex

1= ∞ .

NOTES ON INFINITE SEQUENCES AND SERIES 3

1.6. Bounded Monotonic Sequences. A monotonic sequence is asequence that always increases or always decreases. For instance, 1/n isa monotonic decreasing sequence, and n = 1, 2, 3, 4, . . . is a monotonicincreasing sequence.

A sequence is bounded if its terms never get larger in absolutevalue than some given constant. For instance 1/n is bounded, because|1/n| < 2 for every n, but n is unbounded.

A property of any bounded (increasing or decreasing) sequence isthat it always has a (finite) limit. This might not seem very useful ifwhat we want is to actually compute the limit, but in some cases itmay help. For instance, consider the sequence defined recursively inthe following way:

a1 =√

6 ,

an+1 =√

6 + an (n ≥ 1) .

It can be shown that an it is a bounded monotonic increasing sequence,1

hence it has some limit A:

limn→∞

an = A .

Now, taking limits on both sides of an+1 =√

6 + an (here we use thatthe sequence has a limit) we get that A =

√6 + A, i.e., A2−A−6 = 0.

Solving this second degree equation we get A = 3 or A = −2. Sincethe sequence is positive, the limit cannot be negative, hence it must beA = 3:

limn→∞

an = 3 .

1.7. Homework Problems. E&P,2 11.2: 9–42, 54, 56.

2. Infinite Series

2.1. Series. An infinite series is an expression of the form∞∑

n=1

an = a1 + a2 + a3 + · · · ,

1Proof: We use induction. First note that 0 < a1 =√

6 < 3. By adding 6 andtaking square roots we get

√6 <

√6 + a1 <

√6 + 3 = 3, i.e.: a1 < a2 < 3. Now

assume an < an+1 < 3 for a given n ≥ 1 (induction hypothesis). Again, by adding6 and taking square roots we get

√6 + an <

√6 + an+1 < 3, i.e. an+1 < an+2 < 3

(induction step). From here we get that an < an+1 < 3 for every n ≥ 1, whichproves both, an is increasing and is bounded by 3.

2C.H. Edwards, Jr. & David E. Penney: Calculus with Analytic Geometry, 5thedition, Prentice Hall.

4 MIGUEL A. LERMA

where {an} is a sequence of numbers—sometimes the series starts atn = 0 or some other term instead of n = 1. Its Nth partial sum is

SN =

N∑n=1

an = a1 + a2 + a3 + · · ·+ aN .

2.2. Sum of a Series. The sum

S =∞∑

n=1

an

of a series is defined as the limit of its partial sums

S = limN→∞

SN = limN→∞

N∑n=1

an

if it exists—it this case we say that the series converges. For instance,consider the following series:

∞∑n=1

1

2n=

1

2+

1

4+

1

8+ · · ·

Its partial sum is

SN =

N∑n=1

1

2n=

1

2+

1

4+

1

8+ · · · 1

2N= 1− 1

2N.

Hence, its sum is

S = limN→∞

SN = limN→∞

(1− 1

2N

)= 1 ,

i.e.:∞∑

n=1

1

2n= 1 .

A series may or may not have a sum. For instance, in the followingseries:

∞∑n=0

(−1)n = 1− 1 + 1− 1 + 1− 1 + · · ·

the sequence of partial sums SN = 1, 0, 1, 0, 1, 0, . . . diverges, and theseries has no sum.


2.3. Telescopic Series. Telescopic series are series for which all termsof its partial sum can be canceled except the first and last ones. Forinstance, consider the following series:

∞∑n=1

1

n (n + 1)=

1

2+

1

6+

1

12+ · · ·

Its nth term can be rewritten in the following way:

an =1

n (n + 1)=

1

n− 1

n + 1.

Hence, its Nth partial sum becomes:

SN =N∑

n=1

1

n (n + 1)=

N∑n=1

(1

n− 1

n + 1

)

=

(1− 1

2

)+

(1

2− 1

3

)+

(1

3− 1

4

)+ · · ·

(1

N− 1

N + 1

)

= 1− 1

N + 1.

Hence:∞∑

n=1

1

n (n + 1)= lim

N→∞

(1− 1

N + 1

)= 1 .

2.4. Geometric series. A geometric series

∞∑n=0

an is a series in which

each term is a fixed multiple of the previous one: an+1 = r an, where ris called the ratio. A geometric series can be rewritten in this way:

∞∑n=0

a rn = a + a r + a r2 + a r3 + · · · .

If |r| < 1 its sum is

∞∑n=0

a rn =a

1− r.

Note that a is the first term of the series. If a 6= 0 and |r| ≥ 1, theseries diverges.

Examples:

∑n=0

1

2n= 1 +

1

2+

1

4+

1

8+ · · · = 1

1− 12

= 2 .

6 MIGUEL A. LERMA

∑n=0

(−1)n

2n= 1− 1

2+

1

4− 1

8+ · · · = 1

1− (−12)

=2

3.

Note that in the last example r = an+1/an = (−1)n+1/2n+1

(−1)n/2n = −1/2.

2.5. Termwise addition and multiplication of series. If

∞∑n=1

an = A

and

∞∑n=1

an = B then

∞∑n=1

(an ± bn) =∞∑

n=1

an +∞∑

n=1

bn = A± B

and∞∑

n=1

c an = c

∞∑n=1

an = c A ,

where c is a constant.

2.6. The nth Term Test for Divergence. If an does not converge

to 0, then∞∑

n=1

an does not converge—Note: the reciprocal is not true

in general! (a counterexample is the harmonic series; see below.)

For instance, the series

∞∑n=1

sin n does not converge because sin n does

not converge to 0.

2.7. The Harmonic Series. The series∞∑

n=1

1

n= 1 +

1

2+

1

3+

1

4+ · · ·

is called harmonic series.The harmonic series diverges. This can be proven graphically, by

looking at the graph of the function f(x) = 1/x (fig. 1). The termsof the harmonic series are the areas of the rectangles. Their sum isgreater than the area under the graph of f(x) = 1/x between 1 and∞, which can be computed with the following improper integral:∫ ∞

1

1

xdx = lim

M→∞

∫ M

1

1

xdx = lim

M→∞[ln x]M1 = lim

M→∞(ln M − ln 1) = ∞ .


1/41/31/21

y=1/x

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1 2 3 4 5x

Figure 1. The harmonic series

Hence,∞∑

n=1

1

n= ∞ .

2.8. Series that are Eventually the Same. If an = bn for every n

large enough, then the series

∞∑n=1

an and

∞∑n=1

bn either both converge or

both diverge. In other words, the convergence or divergence of a series

depends only on its “tail”

∞∑n=k

an.

2.9. Homework. E&P, 11.3: 1–37, 49, 50, 64. Maple Worksheet:WS1 (due 10/7/99).

3. Taylor Series and Taylor Polynomials

3.1. Linear Approximation of a Function. Assume that we wantto compute the value of a function such as sin 0.1 or ln 1.1. Evaluatingpolynomials is easy, it can be accomplished by a sequence of compu-tations involving only arithmetic operations (+,−,×). For instance, iff(x) = x2 +7x+2 then f(3) = 32 +7×3+2 = 32. For other functionsin general we may not be able to evaluate them exactly, but we can

8 MIGUEL A. LERMA

do it approximately. A first approximation consists of substituting thegraph of the function by a tangent line (fig. 2).

(a,f(a))

a

P1(x) = c0 + c1 (x-a) y=f(x)

–0.5

0

0.5

1

0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4x

Figure 2. Linear approximation

The equation of the tangent line at x = a is

P1(x) = c0 + c1(x− a) .

Also, the following conditions must be met:

P1(a) = f(a)

P ′1(a) = f ′(a)

i.e., the tangent line must pass through the point (a, f(a)), and itsslope should be the derivative of f(x) at x = a. From here we get:

c0 = f(a)

c1 = f ′(a)

hence the tangent line and first approximation of f(x) at x = a is

P1(x) = f(a) + f ′(a) (x− a) .

P1(x) is called the 1st-degree Taylor polynomial of f(x) at x = a.For instance, if f(x) = ln x then its first-degree Taylor polynomial

at x = 1 is

P1(x) = ln 1 +1

1(x− 1) = x− 1 .


Hence

ln x ≈ x− 1

for x close to 1. In particular ln 1.1 ≈ 1.1− 1 = 0.1. Compare to theactual value ln 1.1 = 0.095310179 . . . .

3.2. Higher Degree Polynomial Approximations. The linear (firstdegree) polynomial approximation might be enough for many practicalpurposes, but sometimes we need a better approximation. This can beaccomplished by using higher degree polynomials.

The nth degree Taylor polynomial of a function f(x) at x = a is apolynomial of the form:

Pn(x) = c0 + c1 (x− a) + c2 (x− a)2 + · · ·+ cn (x− a)n

=n∑

k=0

ck (x− a)k

whose value and derivatives up to the nth are equal to those of f(x)at x = a, i.e.:

Pn(a) = f(a)

P ′n(a) = f ′(a)

P ′′n (a) = f ′′(a)

. . .

P (nn (a) = f (n(a)

Solving those equations we get that ck = f (k(a)/k!, hence the nth-degree Taylor polynomial of f(x) at x = a is

Pn(x) =f(a)

0!+

f ′(a)

1!(x− a) +

f ′′(a)

2!(x− a)2 + · · ·+ f (n(a)

n!(x− a)n

=

n∑k=0

f (k(a)

k!(x− a)k .

For instance, for f(x) = ln(x) and a = 1 we get

P2(x) = (x− 1)− 1

2(x− 1)2 ,

hence, ln 1.1 ≈ (1.1−1)− 12(1.1−1)2 = 0.095, closer to the actual value

ln 1, 1 = 0.095310179 . . . than the first degree approximation computedabove.

10 MIGUEL A. LERMA

3.3. Taylor series. The next logical step is to use an “infinite”-degreepolynomial, i.e., a series of the form:

∞∑k=0

f (k(a)

k!(x− a)k .

This series is called the Taylor series of f(x) at x = a. If a = 0 thenthe Taylor series is called Maclaurin series.

A function f(x) such that

1. its Taylor series converges, and

2. f(x) =

∞∑k=0

f (k(a)

k!(x− a)k ,

is called analytic. A few examples of analytic functions are the follow-ing:

ex = 1 + x +x2

2!+

x3

3!+ · · · =

∞∑n=0

xn

n!

ln (1 + x) = x− x2

2+

x3

3− · · · =

∞∑n=1

(−1)n+1 xn

n(|x| < 1)

sin x = x− x3

3!− x5

5!+ · · · =

∞∑n=0

(−1)n x2n+1

(2n + 1)!

cos x = 1− x2

2!+

x4

4!− · · · =

∞∑n=0

(−1)n x2n

(2n)!

1

1 + x= 1− x + x2 − x3 + · · · =

∞∑n=0

(−1)nxn (|x| < 1)

(1 + x)α =

(α

0

)+

(α

1

)x +

(α

2

)x2 +

(α

3

)x3 + · · · =

∞∑n=0

(α

n

)xn

(|x| < 1)

where

(α

n

)=

α (α− 1) (α− 2)(n factors). . . (α− n + 1)

n!.

As an example, the Taylor series for ex can be used for computingthe number e:

e = e1 =∞∑

n=0

1

n!= 1 + 1 +

1

2!+

1

3!+ · · · = 2.718281828 . . .

3.4. Homework. E&P, 11.4: 11–20, 23–25, 29–40.


4. The Integral Test.

In section 2.7 we found that the harmonic series diverges by com-paring its sum with an integral (see fig. 1). This technique is calledintegral test. More generally we have that the following holds:

4.1. Integral Test. Suppose∑∞

n=1 an is a positive-term series (i.e.,an > 0 for every n = 1, 2, 3, . . . ). Also assume that f(x) is a positivevalued, decreasing, continuous function for x ≥ 1 such that an = f(n)

for every n = 1, 2, 3 . . . . Then the series∞∑

n=1

an and the improper

integral

∫ ∞

1

f(x) dx either both converge or both diverge.

So, in order to test the series for convergence it is enough to test thecorresponding integral for convergence, which in many cases is easier.

The next one is a typical example of application of the integral test.

4.2. p-Series. A p-series is a series of the form:∞∑

n=1

1

np,

where p > 0 is a fix exponent.The case p = 1 is the harmonic series, which was shown to be diver-

gent in section 2.7. On the other hand, if p 6= 1:∫ M

1

1

xpdx =

[x1−p

1− p

]M

1

=1

1− p

(M1−p − 1

).

As M → ∞ the limit of the above expression converges for p > 1,and diverges for p ≤ 1. Hence, the integral test shows that the p-seriesconverges for p > 1, and diverges for p ≤ 1.

4.3. Homework. E&P, 11.5: 1–30, 35–38.

5. Comparison Test for Positive-Term Series

The comparison test allows us to test a series for convergence bycomparing it to another series for which convergence is easier to test.It says the following:

5.1. Comparison Test. Suppose∑

an and∑

bn are positive-termseries such that an ≤ bn for every n. Then:

1. If∑

an diverges then∑

bn diverges.2. If

∑bn converges then

∑an converges.

12 MIGUEL A. LERMA

For instance, consider the series∞∑

n=1

1√n3 + 1

.

Since 1/√

n3 + 1 < 1/√

n3 = 1/n3/2, the convergence of the given seriescan be derived from the convergence of the following series:

∞∑n=1

1

n3/2.

This is a p-series with p = 3/2 > 1, hence it converges, and so does thegiven series.

In some cases the ordinary comparison test is hard to apply, but thefollowing is easier to use:

5.2. Limit Comparison Test. Suppose∑

an and∑

bn are positive-term series, and the limit

L = limn→∞

an

bn

exists and is not zero nor infinity: 0 < L < +∞. Then either bothseries converge or both diverge.

For instance, look at the following series:∞∑

n=1

2 +√

3n√n3 + n + 1

.

We can test its convergence by limit comparison with the harmonicseries:

∞∑n=1

1

n.

In fact:

limn→∞

(2 +√

3n)/√

n3 + n + 1

1/n=√

3 ,

which is not zero nor infinity. Since the harmonic series diverges, weconclude that the given series also diverges.

5.3. Rearrangement and grouping. In a positive-term series rear-ranging and regrouping its terms does not alter its sum. Note that thisis not true in general for other series. For instance, the following seriesis divergent:

1− 1 + 1− 1 + 1− 1 + · · · .


However, if we regroup its terms in the following way, we get a conver-gent series with zero sum:

(1− 1) + (1− 1) + (1− 1) + · · · = 0 + 0 + 0 + · · · = 0 .

A different regrouping still give us a different sum:

1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + · · · = 1 + 0 + 0 + 0 + · · · = 1 .

However, if the terms are all positive, then we can rearrange andregroup them without any fear of changing the sum. So, for instance:

1

2+

1

4+

1

8+

1

16+ · · · =

∑n=1

1

2n=

1/2

1− 1/2= 1 .

Now, if we regroup its terms in the following way, we get:(1

2+

1

4

)+

(1

8+

1

16

)+ · · · = 3

4+

3

16+ · · · =

∑n=1

31

4n=

3/4

1− 1/4= 1 ,

so, we get the same sum.

5.4. Homework. E&P, 11.6: 1–36.

6. Alternating Series and Absolute Convergence

An alternating series is a series of the form

∞∑n=1

(−1)n+1an = a1 − a2 + a3 − a4 + · · ·

or∞∑

n=1

(−1)nan = −a1 + a2 − a3 + a4 − · · ·

where an > 0 for every n. Examples: the alternating harmonic series:

∞∑n=1

(−1)n+1 1

n= 1− 1

2+

1

3− 1

4+ · · · ;

an alternating geometric series:

∞∑n=0

(−1)n 1

2n= 1− 1

2+

1

4− 1

8+ · · · .

14 MIGUEL A. LERMA

6.1. Alternating Series Test. If an alternating series verifies:

1. an it is decreasing: an ≥ an > 0 for every n, and2. the nth term tends to zero: limn→∞ an = 0 ,

then the series converges.So, in this particular case the “reciprocal” of the nth term test holds.E.g.:

∞∑n=1

(−1)n+1

n= 1− 1

2+

1

3− 1

4+ · · · = ln 2

∞∑n=0

(−1)n+1

2n + 1= 1− 1

3+

1

5− 1

7+ · · · = π

4.

6.2. Alternating Series Remainder Estimate. Let S be the sumof an alternating series:

S =

∞∑n=0

(−1)n+1an ,

and SN its Nth partial sum:

SN =

N∑n=0

(−1)n+1an .

Then the difference RN = S − SN (remainder) between the sum ofthe series and that of the Nth partial sum has the same sign as thefollowing term of the series (−1)n+2 an+1, and

0 ≤ |RN | < an+1 .

In particular, S is always between two consecutive partial sums.For instance, we know that

π = 4

∞∑n=0

(−1)n+1

2n + 1,

but how close do we get to π by adding, say, one hundred terms of thatseries? Answer: its sum π is between

499∑

n=0

(−1)n+1

2n + 1= 3.131592904 . . .

and

4100∑n=0

(−1)n+1

2n + 1= 3.151493401 . . .


Compare to the actual value π = 3.141592654 . . . .

6.3. Absolute Convergence. A general series∑

an is said to be ab-solutely convergent if the series of absolute values of its terms

∑ |an|is convergent.

We have that a series can be:

1. Convergent and absolutely convergent, e.g:

∞∑n=0

(−1

2

)n

= 1− 1

2+

1

4− 1

8+ · · ·

2. Convergent but not absolutely convergent—in this case the seriesis called conditionally convergent—, e.g:

∞∑n=1

(−1)n+1

n= 1− 1

2+

1

3− 1

4+ · · ·

3. Not convergent nor absolutely convergent, e.g:

∞∑n=1

n = 1 + 2 + 3 + 4 + · · ·

However, a series cannot be absolutely convergent and not conver-gent, because absolute convergence implies convergence:

absolute convergent =⇒ convergent

Example: Does the series

∞∑n=1

cos n

n2converge? Answer: Look at the

series of absolute values:∞∑

n=1

| cos n|n2

≤∞∑

n=1

1

n2.

By comparison test, it converges (the right hand side is a p-series withp > 1), hence the given series is absolutely convergent, which impliesthat it is indeed convergent.

6.4. Ratio Test. Suppose that the limit ρ = limn→∞

∣∣∣∣an+1

an

∣∣∣∣ exists or is

infinity. Then

1. If ρ < 1 =⇒ ∑an converges absolutely.

2. If ρ > 1 =⇒ ∑an diverges.

3. If ρ = 1 =⇒ the ratio test is inconclusive.

16 MIGUEL A. LERMA

As a rule of thumb, for geometric series ρ = |r| (the ratio), and theconclusion of the ratio test is analogous to the one for geometric series,i.e., the series converges for |r| < 1 and diverges for |r| > 1.

Example: For the series

∞∑n=1

n2

2nwe have

ρ = limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

(n + 1)2/2n+1

n2/2n= lim

n→∞(n + 1)2

2 n2=

1

2< 1 ,

hence, it converges absolutely.3

6.5. Root Test. In some cases in which the ratio test is unable to pro-vide an answer, the root test may help. It says the following: Suppose

that the limit ρ = limn→∞

n√|an| exists or is infinity. Then

1. If ρ < 1 =⇒ ∑an converges absolutely.

2. If ρ > 1 =⇒ ∑an diverges.

3. If ρ = 1 =⇒ the root test is inconclusive.

Example: Consider the following series∞∑

n=1

1

2n+sinn. For this series

the ratio test cannot be used, becausean+1

an= 2−1+sinn−sin (n+1) = 2−1−2 sin 1

2cos (n+ 1

2)

which has no limit. However, the root tests shows that the series isabsolutely convergent:

limn→∞

n

√1

2n+sinn= lim

n→∞1

21+sin n/n=

1

2< 1 .

6.6. Homework. E&P, 11.7: 1–42.

7. Power Series

A power series is a sort of infinite polynomial of the form∞∑

n=0

an xn

Taylor series are particular cases of power series. E.g.:

ex =∞∑

n=0

xn

n!= 1 + x +

x2

2!+

x3

3!+ · · ·

3The sum is exactly∞∑

n=1

n2

2n= 6.


sin x =∞∑

n=0

(−1)n x2n+1

(2n + 1)!= x− x3

3!− x5

5!+ · · ·

cos x =∞∑

n=0

(−1)n x2n

(2n)!= 1− x2

2!+

x4

4!− · · ·

1

1 + x=

∞∑n=0

(−1)nxn = 1− x + x2 − x3 + · · · (|x| < 1)

7.1. Convergence of a Power Series. The first question we mustanswer about a powers series is “for what values of x does the seriesconverge?” This can be answered with the ratio test:

ρ = limn→∞

∣∣∣∣an+1xn+1

anxn

∣∣∣∣ = |x| limn→∞

∣∣∣∣an+1

an

∣∣∣∣ =|x|R

,

where

1

R= lim

n→∞

∣∣∣∣an+1

an

∣∣∣∣ .

(If the limit is zero then we take R = ∞. If it is infinity then R = 0.)

We know that the series converges absolutely if ρ < 1, i.e., |x|R

< 1,

and diverges if ρ > 1, i.e., |x|R

> 1. Hence the series converges absolutelyfor |x| < R, i.e., on the interval (−R, R), and diverges for |x| > R. Thenumber R is called the radius of convergence of the series.

It remains to determine if the series converges at the endpoints x = Rand x = −R. This can be answered by substituting x = R and x = −Rin the power series and studying the resulting series. After doing that,we will find that the interval of convergence is one of the following:

(−R, R) , [−R, R) , (−R, R] or [−R, R]

If R = ∞ then the interval of convergence is the whole real line R.Example: consider the Maclaurin series for ln(1 + x):

∞∑n=1

(−1)n+1 xn

n.

Using the ratio test we get:

ρ = limn→∞

∣∣∣∣xn+1/(n + 1)

xn/n

∣∣∣∣ = limn→∞

|x| n

n + 1= |x| .

Hence the series converges absolutely for |x| < 1 and diverges for|x| > 1. It remains to study the endpoints x = ±1.

18 MIGUEL A. LERMA

For x = 1 the power series becomes

∞∑n=1

(−1)n+1 1

n,

which is the alternating harmonic series. We know that it converges.For x = −1 the power series becomes

−∞∑

n=1

1

n,

which is the (usual) harmonic series. We know that it diverges.Hence the interval of convergence is (−1, 1].

7.2. Power Series in Powers of (x− c). The same techniques canbe applied to power series of the form:

∞∑n=0

an (x− c)n

Using the ratio test we get

ρ = limn→∞

∣∣∣∣an+1(x− c)n+1

an(x− c)n

∣∣∣∣ = |x− c| limn→∞

∣∣∣∣an+1

an

∣∣∣∣ =|x− c|

R,

where

1

R= lim

n→∞

∣∣∣∣an+1

an

∣∣∣∣ .

Hence the series converges absolutely if |x−c|R

< 1 and diverges if|x−c|

R> 1. In other words, it converges absolutely for |x− c| < R, i.e.,

on the interval (c−R, c + R), and diverges for |x− c| > R. As above,the endpoints x = c− R and x = c + R must be tested separately.

So the radius of convergence is as before, but the interval of conver-gence is centered at x = c instead of x = 0.

7.3. The Binomial Series. The binomial series is the Maclaurin se-ries of the function f(x) = (1 + x)α. It can be computed in the usualway:

f(x) = (1 + x)α f(0) = 1f ′(x) = α (1 + x)(α−1) f ′(0) = αf ′′(x) = α (α− 1) (1 + x)(α−2) f ′′(0) = α (α− 1)

. . . . . .


f (n(x) = α (α− 1) . . . (α− n + 1) (1 + x)(α−2)

. . .

f (n(0) = α (α− 1) . . . (α− n). . .

Hence:

(1 + x)α =

(α

0

)+

(α

1

)x +

(α

2

)x2 +

(α

3

)x3 + · · · =

∞∑n=0

(α

n

)xn

(|x| < 1)

where

(α

n

)=

α (α− 1) (α− 2)(n factors). . . (α− n + 1)

n!.

Note that if α = m a positive integer, then(

mn

)= 0 for n > m, so

the series becomes a finite sum identical to the binomial expansion:

(1 + x)m =

m∑n=0

(m

n

)xn

Other examples:1

1 + x= (1 + x)−1 =

∞∑n=0

(−1)nxn = 1− x + x2 − x3 + · · ·√

1 + x = (1 + x)1/2 =

∞∑n=0

(1/2

n

)xn = 1 +

1

2x− 1

8x2 +

1

16x3 + · · ·

1√1 + x

= (1 + x)−1/2 =∞∑

n=0

(−1/2

n

)xn = 1− 1

2x +

3

8x2 − 5

16x3 + · · ·

7.4. Differentiation and Integration of Power Series. If a func-tion f(x) can be represented as a power series:

f(x) =∞∑

n=0

an xn

with radius of convergence R, then:

1. It is differentiable on (−R, R) and

f ′(x) =∞∑

n=1

n an xn−1 .

2. It is integrable on (−R, R) and∫ x

0

f(t) dt =

∞∑n=0

an xn+1

n + 1.

20 MIGUEL A. LERMA

As an application, we use this result for computing the power seriesfor the arctangent f(x) = tan−1 x. In fact, we have:

f ′(t) =1

1 + t2= 1− t2 + t4 − t6 + · · · =

∞∑n=0

(−1)nt2n .

Integrating termwise we get:

f(x) =

∫ x

0

f ′(t) dt = x− x3

3+

x5

5− x7

7+ · · · =

∞∑n=0

(−1)n t2n+1

2n + 1.

Hence:

tan−1 x =

∞∑n=0

(−1)n t2n+1

2n + 1= x− x3

3+

x5

5− x7

7+ . . . .

7.5. Homework. E&P, 11.8: 1–12. Maple Worksheet: WS2 (due10/14/99).

8. Power Series Computations

8.1. Adding and Multiplying Series.

If f(x) =∞∑

n=0

an xn and g(x) =∞∑

n=0

bn xn then

f(x) + g(x) =∞∑

n=0

(an + bn) xn

and

f(x) g(x) =

∞∑n=0

cn xn

where

cn = a0 bn + a1 bn−1 + · · ·an b0 =n∑

k=0

ak bn−k

As an example, we can compute the first few terms of the powerseries for tanx = a0 + a1 x + a2 x2 + · · · , using the power series forsin x = x−x3/3!+x5/5!−· · · and cos x = 1−x2/2!+x4/4!−· · · , andthe relation sin x = tan x cos x:

sinx︷︸︸︷x− x3

3!+

x5

5!− · · · =

tan x︷︸︸︷(a0 + a1 x + a2 x2 + a3 x3 + · · · )

cos x︷︸︸︷(1− x2

2!+

x4

4!− · · ·

)

= a0 + a1 x +(−a0

2+ a2

)x2 +

(−a1

2+ a3

)x3 + · · ·


Identifying coefficients we get:

a0 = 0a1 = 1

−12a0 +a2 = 0

−12a1 +a3 = −1

6. . .

Solving that system of equations we get a0 = 0, a1 = 1, a2 = 0, a3 =1/3, . . . , hence:

tanx = x +x3

3+ · · ·

8.2. Computing Limits with Power Series. Since power series arecontinuous in their interval of convergence, we have that if

f(x) =∞∑

n=0

an (x− c)n = a0 + a1 x + a2 x2 + · · ·

then

limx→c

f(x) = f(c) = a0 + 0 + 0 + · · · = a0 .

This can be applied to computing limits of indeterminate forms f(x)/g(x)by substituting power series for f(x) and g(x). Example:

limx→0

1− cos x

x sin x= lim

x→0

1− (1− x2/2 + x4/24− · · · )x (x− x3/6 + · · · )

= limx→0

x2/2− x4/24 + · · ·x2 − x4/6 + · · ·

= limx→0

1/2− x2/24 + · · ·1− x2/6 + · · ·

=1

2

8.3. Homework. E&P, 11.9: 23–28.

NOTES ON INFINITE SEQUENCES AND SERIESmlerma/courses/b17-99f/seq.pdf · NOTES ON INFINITE SEQUENCES AND SERIES 5 2.3. Telescopic Series. Telescopic series areseries forwhich allterms

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